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The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A.
B.
C.
D.
The given equations are
x2 + y2 = 16 … (1)
y2 = 6x … (2)
Area bounded by the circle and parabola
=2[area(OADO)+area(ADBA)]=2[∫206x‾‾√ dx+∫4216−x2‾‾‾‾‾‾‾√ dx]=2∫206x‾‾√ dx+2∫4216−x2‾‾‾‾‾‾‾√ dx=26‾√∫20x√ dx+2∫4216−x2‾‾‾‾‾‾‾√ dx=26‾√×23[x32]20+2[x216−x2‾‾‾‾‾‾‾√+162sin−1(x4)]42 =46√3(22‾√−0)+2[{0+8sin−1(1)}−{23‾√+8sin−1(12)}]=163√3+2[8×π2−23‾√−8×π6]=163√3+2(4π−23‾√−4π3)=163√3+8π−43‾√−8π3=163√+24π−43√−8π3=16π+123√3=43[4π+3‾√] square units=2areaOADO+areaADBA=2∫026x dx+∫2416-x2 dx=2∫026x dx+2∫2416-x2 dx=26∫02x dx+2∫2416-x2 dx=26×23x3202+2x216-x2+162sin-1x424 =46322-0+20+8sin-11-23+8sin-112=1633+28×π2-23-8×π6=1633+24π-23-4π3=1633+8π-43-8π3=163+24π-43-8π3=16π+1233=434π+3 square units
Area of circle = π (r)2
= π (4)2
= 16π square units
∴ Required area=16π−43(4π+3‾√)=16π−16π3−43√3=32π3−43√3=43[8π−3‾√] square units∴ Required area=16π-434π+3=16π-16π3-433=32π3-433=438π-3 square units
Thus, the correct answer is C.
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